Motor Overheating w/ Speed

I got something to clear up inside my head about DC motors. This all has to do with how the speed of a motor *seems* to have an affect on a motor overheating.

Motor heat dissipation is a result of the I^2xR losses inside the motor and the current draw of a DC motor is only dependent on the torque and independent of speed. THis means that motor overheating is only a result of the torque output and not the speed right?

Now, If a motor was made from insulation materials of infinite dielectric strength and zero friction, you could get a motor to produce any torque at any speed by gearing down by however much was required and then hiking up the voltage by any amount to speed it up right?

But in real life, this obviously isn't the case. What we have to do if we need more speed is reduce the gear ratio in order to increase output speed. But doing this also increases the torque requirements on the motor. So while it appears that trying to increase the output speed causes the motor to overheat, the actual cause is how the reduction in gear ratios increases torque requirements on the motor...
Right?
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This is what I need this for:
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You could gear a motor to move a robot quickly on a flat surface where the torque requirements are low and the output speed is high, and the total power output is say, 100W. But now if that same vehicle tried to move up an incline that requires increased torque, but greatly reduced speed, where the power works out to be 20W that doesn't mean that the aformentioned setup won't be overheat the motor, right? Since the torque required is much higher, the current is much higher too which causes the motor to overload, even though the power output is less.

So I should not be sizing the motors by their power rating for movement on two different types of terrain so much as their torque rating correct? Since, for a given gear ratio, the motor could output more power just fine, while in another situtation the power output could be much less but the motor would still overheat.

The other main source of heat in a DC motor is the commutator. I once rewound a RS540 motor to work on 30V and it worked fine except at high speed when the solder on the commutator connections melted!!

I think a useful exercise is to think of what happens if you connect a motor to a constant current source. In your example above, on flat ground the motor will increase in speed until frictional losses are equal to torque or the motor flies apart. On the incline the motor will slow down until the torque produced is equal to friction plus the work done raising the weight. In both cases the I²R losses are the same and the only additional heat is due to friction. A ballraced brushless motor would probably reach the same temperature in both cases.

How did you arrive at your figures? I would expect a motor to use more power on an incline than traveling level.

Mike.

It's only the same amount of power if you are going the same speed on both the incline and the flat ground. Usually the incline isn't a smooth ramp so you might slow down by 10x but the torque is only increased by 3x so you end up using less power.

7.5ft/s on a plane vs. 1ft/s up a 45 degree incline, for example.

I was sizing my motor by power ratings for travelling at nominal conditions of 7.5ft/s on a 10 degree incline with 0.3 friction which uses something like 265W. But the peak torque conditions is 1ft/s up a 70 degree incline with 0.5 friction. The peak torque condition uses only 80W but draws a 225% more current which I will overheat the motor.

I don't have real graphs, I just have 3 performance points (voltage, current, RPM, and power input) to extrapolate from so I don't even have the torque constants or the actual efficiency. I've had to assume a bunch of stuff so I don't know for sure how much current it will actually draw. I've had to make adjust all my torque, speed and gear ratio calculations "fit" to one of the 3 points.

EDIT: It just occured to me that I can take the extrapolating one more step. If I fit the gear ratio calculations to a particular performance point I have the current at that operating point. So presumably if I change the gear ratio then I also scale that current at that performance by the same amount. I don't know if it's true that the speed will also scale the same way. But if I gear it up it would probably push it closer to the maximum efficiency point anyways since speed isn't really an issue as much as the distance it can travel on a given battery charge.

If at 7.5ft/s on a 10 degree incline (+30%) you require 265W then your vehicle must weigh around 110lb. That same vehicle at 1ft/s on a 70 degree incline (+50%) would require 220W.

Ignoring friction completely,
1*sin(70)/(7.5*sin(10)) = 0.72

Your second figure should be at least 72% of your first = 265*0.72 = 191W.

One of us has a mistake in their calculations. It may be me.

Mike.
Edit, As long as you have stall torque and no load RPM then you can produce a power curve graph. See link.

I assumed 60% motor efficiency and 60% efficiency of the drivetrain which approximately doubles the power required. Also, the friction of 0.3 (as compared to zero) hugely affects the power calculation- it seems to double or even triple the power required. So there is about a difference of a factor of 4 which is about the same scaling of 110lbs vs 20lbs.

I'm just really conservative on these ratings because I had to guess the efficiency of the gearbox- 4 spur gears and 2 wishbone U-joints. The friction is also conservative since the tires are these foam filled things that deform and the ground is bumpy and all.

Unfortunately, I do not have stall torque.

Why doesn't a friction value of 0.3 simply increase power requirement by 30%?

Do you have the DC resistance of the motors or better still a link?

Mike.

https://www.aircraft-world.com/prod_datasheets/hp/z30/z3025spec.htm

The 12-turn version. The friction value of 0.3 doesn't increase the power requirement by 30% because some of the weight is taken off the wheels due to the incline. I'd attach the excel file if I could it seems that I cannot.

THis, however, is the formula that I am using to calculate the force required by the wheel:
=<weight>*sin<inline> + <friction>*<weight>*cos(incline)

I basically did what I wrote in my tutorial:
https://www.electro-tech-online.com/threads/motor-sizing-for-moving-robots.23264/

If I ever decide to dish out the money for all the parts, I think there's a good chance it will end up better than what I calculated, particularily the efficiency of the drivetrain might be about 10% better and the coefficient of friction might be half of what I guessed. THere's also going to be a front and rear motor so power while travelling on ground where at least one front and one rear wheels are in firm contact with the ground won't be a problem. But I'm sizing peak conditions for when the thing ends up hanging off of a ledge or something by just the front wheels.

At higher loads your IR losses rise very rapidly. 60% is a good assumption only at approximately (RPMmax/2).

Where did you come up with the 30% value. I think you are out by a factor of 10. I refer to wiki, the value on the page of 300N to move a 1000kg car is 3%. If the value was 30% then that would mean my car would not roll down a hill of less than 28°.

Mike.

I pulled the friction of 0.3 out of the air. It's because the wheels have these giant nubs that can deform and are also foam filled and the foam is likely not stiff enough to maintain a "completely" round shape under the weight.

My current values being used for nominal conditions and peak conditions are:
----------------Nominal----PEak

Weight (lbs)..........20.00.....20.00
Incline (Degrees) 10.00......45.00
Friction................0.20.......0.20
Force (lbs)............7.412194565........16.97056275
Torque (lbs-in).......29.64877826......67.88225099

Speed (Feet/s)......9.50........1.00
Reduction..............26..........26
RPM..............7079.617834.....745.2229299

Motor Efficiency.....0.6.................0.6
Gearbox Efficiency.........0.6........0.6
Power Output (W)......95.59......23.04
Power Input (W).......265.5264188....63.99316143

From my RC car experience: If the motor overheats, it is geared too high. Change the pinion gear to a lower value. This will of course allow the motor to get more rpms, but this always solved the problem for us. We were using 8 to 29 turn motors, mostly double winds.

As the motor rpm decreases it will draw more current, right?